Problem: $\int^{e}_{1}\dfrac{(\ln(x))^2}{x}\,dx\, = $
Explanation: Strategy Let's first find the indefinite integral $\int \dfrac{(\ln(x))^2}{x}\,dx\,$. Then we can use the result to compute the definite integral. Finding the indefinite integral To find $\int \dfrac{(\ln(x))^2}{x}\,dx\,$, we can use U-substitution. If we let $ {u=\ln(x)}$, then ${du=\dfrac1x \, dx}$. So we have: $\begin{aligned}\int \dfrac{(\ln(x))^2}{x}\,dx\,&=\int {({\ln(x)})^2}\cdot {\dfrac1{x}\,dx}\,\,\\\\\\\\ &=\int u^2 {\, du}\,\\\\\\\\ &=\dfrac{1}{3} u^3+C\\\\\\\\ &=\dfrac{1}{3} (\ln(x))^3+C \end{aligned}$ Evaluating the definite integral Now let's compute the definite integral: $\begin{aligned}\int^{e}_{1}\dfrac{(\ln(x))^2}{x}\,dx\, &=\dfrac{1}{3} (\ln(x))^3\Bigg|^e_1\\\\\\\\ &=\dfrac{1}{3}\left((\ln(e))^3-(\ln(1))^3\right)\\\\\\\\ &=\dfrac13(1-0)\\\\\\\\ &=\dfrac13\end{aligned}$ [Did we have to find the indefinite integral first?] The answer $\int^{e}_{1}\dfrac{(\ln(x))^2}{x}\,dx\, = \dfrac13$